AS-Level Maths:
Core 1
for Edexcel
C1.2 Algebra and
functions 2
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Quadratic expressions
Quadratic expressions
Contents
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Quadratic expressions
A quadratic expression is an expression in which the
highest power of the variable is 2. For example:
x2 – 2
w2 + 3w + 1
4 – 5g2
t2
2
The general form of a quadratic expression in x is:
ax2 + bx + c
(where a ≠ 0)
x is a variable.
a is the coefficient of x2.
b is the coefficient of x.
c is a constant term.
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Factorizing quadratics
Quadratic expressions
Contents
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Factorizing quadratic expressions
Factorizing an expression is the inverse of expanding it.
Expanding or multiplying out
x2 + 3x + 2
(x + 1)(x + 2)
Factorizing
When we expand an expression we multiply out the brackets.
When we factorize an expression we write it with brackets.
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Factorizing quadratic expressions
No constant term
Quadratic expressions of the form ax2 + bx can always be
factorized by taking out the common factor x. For example:
3x2 – 5x = x(3x – 5)
The difference between two squares
When a quadratic has no term in x and the other two terms can
be written as the difference between two squares, we can use
the identity
a2 – b2 = (a + b)(a – b)
to factorize it. For example:
9x2 – 49 = (3x + 7)(3x – 7)
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Factorizing quadratic expressions
Quadratic expressions with a = 1
Quadratic expressions of the form x2 + bx + c can be factorized
if they can be written using brackets as
(x + d)(x + e)
where d and e are integers.
If we expand (x + d)(x + e), we have
(x + d)(x + e) = x2 + dx + ex + de
= x2 + (d + e)x + de
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Factorizing quadratic expressions
The general form
Quadratic expressions of the general form ax2 + bx + c can be
factorized if they can be written using brackets as
(dx + e)(fx + g)
where d, e, f and g are integers.
If we expand (dx + e)(fx + g), we have
(dx + e)(fx + g)= dfx2 + dgx + efx + eg
= dfx2 + (dg + ef)x + eg
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Completing the square
Quadratic expressions
Contents
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Perfect squares
Some quadratic expressions can be written as perfect squares.
For example:
x2 + 2x + 1 = (x + 1)2
x2 – 2x + 1 = (x – 1)2
x2 + 4x + 4 = (x + 2)2
x2 – 4x + 4 = (x – 2)2
x2 + 6x + 9 = (x + 3)2
x2 – 6x + 9 = (x – 3)2
In general:
x2 + 2ax + a2 = (x + a)2
or
x2 – 2ax + a2 = (x – a)2
How could the quadratic expression x2 + 8x
be made into a perfect square?
We could add 16 to it.
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Completing the square
Adding 16 to the expression x2 + 8x to make it into a perfect
square is called completing the square.
We can write
x2 + 8x = x2 + 8x + 16 – 16
If we add 16 we then have to subtract 16 so that both sides
are still equal.
By writing x2 + 8x + 16 we have completed the square and so
we can write this as
x2 + 8x = (x + 4)2 – 16
In general:
2
b

b
x  bx   x     
2

2
2
2
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Completing the square
Complete the square for x2 – 10x.
Compare this expression to (x – 5)2 = x2 – 10x + 25
x2 – 10x = x2 – 10x + 25 – 25
= (x – 5)2 – 25
Complete the square for x2 + 3x.
Compare this expression to ( x + 32 ) 2 = x 2 + 3 x +
2
2
x + 3x = x + 3x +
9
4
= ( x + 32 ) 
9
4
2
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
9
4
9
4
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Completing the square
How can we complete the square for
x2 – 8x + 7?
Look at the coefficient of x.
This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16.
x2 – 8x + 7 = x2 – 8x + 16 – 16 + 7
= (x – 4)2 – 9
In general:
2
2
b

b
x  bx  c   x       c
2

2
2
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Completing the square
Complete the square for x2 + 12x – 5.
Compare this expression to (x + 6)2 = x2 + 12x + 36
x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5
= (x + 6)2 – 41
Complete the square for x2 – 5x + 7.
Compare this expression to ( x  52 ) 2 = x 2  5 x +
x  5x + 7 = x  5x +
2
2
= ( x + 52 ) 
2
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25
4

25
4
25
4
7
3
4
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Completing the square
When the coefficient of x2 is not 1, quadratic equations in the
form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by
completing the square.
Complete the square for 2x2 + 8x + 3.
Take out the coefficient of x2 as a factor from the terms in x:
2x2 + 8x + 3 = 2(x2 + 4x) + 3
By completing the square, x2 + 4x = (x + 2)2 – 4 so
2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3
= 2(x + 2)2 – 8 + 3
= 2(x + 2)2 – 5
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Completing the square
Complete the square for 5 + 6x – 3x2.
Take out the coefficient of x2 as a factor from the terms in x:
5 + 6x – 3x2 = 5 – 3(–2x + x2)
= 5 – 3(x2 – 2x)
By completing the square, x2 – 2x = (x – 1)2 – 1 so
5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1)
= 5 – 3(x – 1)2 + 3
= 8 – 3(x – 1)2
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Complete the square
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Solving quadratic equations
Contents
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Quadratic equations
The general form of a quadratic equation in x is:
ax2 + bx + c = 0 (where a ≠ 0)
Quadratic equations can be solved by:
factorization
completing the square, or
using the quadratic formula.
The solutions to a quadratic equation are called the roots of
the equation.
A quadratic equation may have:
two real distinct roots
one repeated root, or
no real roots.
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The roots of a quadratic equation
If we sketch the graph of a quadratic function y = ax2 + bx + c
the roots of the equation coincide with the points where the
function cuts the x-axis.
As can be seen here, this
can happen twice, once or
not at all.
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Solving quadratic equations by factorization
Solve the equation 5x2 = 3x
Start by rearranging the equation so that the terms are on the
left-hand side:
Don’t divide
5x2 – 3x = 0
through by x!
Factorizing the left-hand side gives us
x(5x – 3) = 0
So
x=0
or
5x – 3 = 0
5x = 3
x=
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3
5
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Solving quadratic equations by factorization
Solve the equation x2 – 5x = –4 by factorization.
Start by rearranging the equation so that the terms are on the
left-hand side.
x2 – 5x + 4 = 0
We need to find two integers that add together to make –5
and multiply together to make 4.
Because 4 is positive and –5 is negative, both the integers
must be negative. These are –1 and –4.
Factorizing the left-hand side gives us
(x – 1)(x – 4) = 0
x–1=0
x–4=0
or
x=1
x=4
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Solving quadratics by completing the square
Quadratic equations that cannot be solved by factorization can
be solved by completing the square.
For example, the quadratic equation
x2 – 4x – 3 = 0
can be solved by completing the square as follows:
(x – 2)2 – 7 = 0
(x – 2)2 = 7
x–2= 7
x=2+ 7
x = 4.65
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or
x=2– 7
x = –0.646 (to 3 s.f.)
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Solving quadratics by completing the square
Solve the equation 2x2 – 4x + 1 = 0 by completing
the square. Write the answer to 3 significant figures.
Start by completing the square for 2x2 – 4x + 1:
2x2 – 4x + 1 = 2(x2 – 2x) + 1
= 2((x – 1)2 – 1) + 1
= 2(x – 1)2 – 2 + 1
= 2(x – 1)2 – 1
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Solving quadratics by completing the square
Now solving the equation 2x2 – 4x + 1 = 0:
2(x – 1)2 – 1 = 0
2(x – 1)2 = 1
( x  1) =
2
1
2
x  1= 
x = 1+
x = 1.71
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1
2
or
1
2
x = 1
1
2
x = 0.293 (to 3 s.f.)
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Using the quadratic equation formula
Any quadratic equation of the form
ax2 + bx + c = 0
can be solved by substituting the values of a, b and c into the
formula
x=
b ±
b  4 ac
2
2a
This formula can be derived by completing the square on the
general form of the quadratic equation.
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Using the quadratic formula
Use the quadratic formula to solve 2x2 + 5x – 1 = 0.
2x2 + 5x – 1 = 0
x=
x=
x=
x=
5 +
4
x = 0.186
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b ±
b  4 ac
2
2a
5 ±
5  (4 × 2 × 1)
2
2×2
5 ±
33
25 + 8
4
or
x=
5 
33
4
x = –2.69 (to 3 s.f.)
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Using the quadratic formula
Use the quadratic formula to solve 9x2 – 12x + 4 = 0.
9x2 – 12x + 4 = 0
x=
x=
x=
x=
b ±
b  4 ac
2
2a
( 1 2 ) ±
(12)  (4 × 9 × 4 )
2
2×9
12 ±
144  144
18
12 ±
0
18
There is one repeated root: x =
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2
3
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Equations that reduce to a quadratic form
Some equations, although not quadratic, can be written in
quadratic form by using a substitution. For example:
Solve the equation t4 – 5t2 + 6 = 0.
This is an example of a quartic equation in t.
Let’s substitute x for t2:
x2 – 5x + 6 = 0
This gives us a quadratic equation that can be solved by
factorization:
(x – 2)(x – 3) = 0
x=2
or
t2 = 2
So
t =  2 or
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x=3
t2 = 3
t= 3
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The discriminant
Contents
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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The discriminant
By solving quadratic equations using the formula
x 
b 
b  4 ac
2
2a
we can see that we can use the expression under the square
root sign, b2 – 4ac, to decide how many roots there are.
When b2 – 4ac > 0, there are two real distinct roots.
When
b2
– 4ac = 0, there is one repeated root: x  
b
.
2a
When b2 – 4ac < 0, there are no real roots.
Also, when b2 – 4ac is a perfect square, the roots of the
equation will be rational and the quadratic will factorize.
b2 – 4ac is called the discriminant of ax2 + bx + c
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The discriminant
We can demonstrate each of these possibilities graphically.
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Graphs of quadratic functions
Contents
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Plotting graphs of quadratic functions
A quadratic function in x can be written in the form:
y = ax2 + bx + c
(where a ≠ 0)
We can plot the graph of a quadratic function using a table of
values. For example:
Plot the graph of y = x2 – 4x + 2 for –1 < x < 5.
x
–1
0
1
2
3
4
5
x2
1
0
1
4
9
16
25
– 4x
+4
+0
–4
–8
– 12
– 16
– 20
+2
y = x2 – 4x + 2
+2
7
+2
2
+2
–1
+2
–2
+2
–1
+2
2
+2
7
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Plotting graphs of quadratic functions
x
–1
0
1
2
3
4
5
y = x2 – 4x + 2
7
2
–1
–2
–1
2
7
The points given in the
table are plotted …
… and the points are then
joined together with a
smooth curve.
The shape of this curve is
called a parabola.
It is characteristic of a
quadratic function.
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y
6
5
4
3
2
1
–1 0
–1
1
2
3
4
5
x
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Parabolas
Parabolas have a
vertical axis of
symmetry …
…and a turning
point called the
vertex.
When the coefficient of x2 is positive the vertex is a minimum
point and the graph is -shaped.
When the coefficient of x2 is negative the vertex is a
maximum point and the graph is -shaped.
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Exploring graphs of the form y = ax2 + bx + c
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Sketching graphs of quadratic functions
When a quadratic function factorizes we can use its factorized
form to find where it crosses the x-axis. For example:
Sketch the graph of the function y = x2 – 2x – 3.
The function crosses the x-axis when y = 0.
x2 – 2x – 3 = 0
(x + 1)(x – 3) = 0
x+1=0
x = –1
or
x–3=0
x=3
The function crosses the x-axis at the points (–1, 0) and (3, 0).
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Sketching graphs of quadratic functions
By putting x = 0 in y = 2x2 – 5x – 3 we can also find where the
function crosses the y-axis.
y = 2(0)2 – 5(0) – 3
y=–3
So the function crosses the y-axis at the point (0, –3).
In general:
The quadratic function y = ax2 + bx + c will
cross the y-axis at the point (0, c).
We now know that the function y = x2 – 2x – 3 passes through
the points (–1, 0), (3, 0) and (0, –3) and so we can place these
points on our sketch.
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Sketching graphs of quadratic functions
We can also use the fact that a
parabola is symmetrical to find the
coordinates of the vertex.
y
(3, 0)
(–1, 0)
0
x
The x coordinate of the vertex is
half-way between –1 and 3.
(0, –3)
(1, –4)
When x = 1,
x=
1+ 3
=1
2
y = (1)2 – 2(1) – 3
y = –4
So the coordinates of the vertex are (1, –4).
We can now sketch the graph.
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Sketching graphs of quadratic functions
In
general:
When a quadratic function is written in the form
y = a(x – α)(x – β), it will cut the x-axis at
the points (α, 0) and (β, 0).
α and β are the roots of the quadratic function.
For example, write the quadratic function y = 3x2 + 4x – 4 in the
form y = a(x – α)(x – β) and hence find the roots of the function.
This function can be factorized as follows,
y = (3x – 2)(x + 2)
It can be written in the form y = a(x – p)(x – q) as
y = 3( x  32 ) ( x   2 )
Therefore, the roots are
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2
3
and  2 .
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Exploring graphs of the form y = a(x – α)(x – β)
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Sketching graphs by completing the square
When a function does not factorize we can write it in completed
square form to find the coordinates of the vertex. For example:
Sketch the graph of y = x2 + 4x – 1 by
writing it in completed square form.
x2 + 4x – 1 = (x + 2)2 – 5
The least value that (x + 2)2 can have is 0 because the square
of a number cannot be negative.
(x + 2)2 ≥ 0
Therefore (x + 2)2 – 5 ≥ – 5
The minimum value of the function y = x2 + 4x – 1 is therefore
y = –5.
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Sketching graphs by completing the square
When y = –5, we have,
(x + 2)2 – 5 = –5
(x + 2)2 = 0
x = –2
The coordinates of the vertex are therefore (–2, –5).
The equation of the axis of
x = –2
y
symmetry is x = –2.
Also, when x = 0 we have
y = x2 + 4x – 1
y = –1
0
x
So the curve cuts the y-axis at
(–1, 0)
the point (–1, 0).
Using symmetry we can now
sketch the graph.
(–2, –5)
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Sketching graphs by completing the square
In general, when the quadratic function y = ax2 + bx + c is
written in completed square form as
a(x + p)2 + q
The coordinates of the vertex will be (–p, q).
The axis of symmetry will have the equation x = –p.
Also:
If a > 0 (–p, q) will be the minimum point.
If a < 0 (–p, q) will be the maximum point.
Plotting the y-intercept, (0, c) will allow the curve to be
sketched using symmetry.
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Exploring graphs of the form y = a(x + p)2 + q
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Examination-style questions
Contents
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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Examination-style question
a) Write 2x2 – 8x + 7 in the form a(x + b)2 + c.
b) Write down the minimum value of f(x) = 2x2 – 8x + 7 and
state the minimum value of x where this occurs.
c) Solve the equation 2x2 – 8x + 7 = 0 leaving your answer in
surd form.
d) Sketch the graph of y = 2x2 – 8x + 7.
2x2 – 8x + 7 = 2(x2 – 4x) + 7
a)
= 2((x – 2)2 – 4) + 7
= 2(x – 2)2 – 8 + 7
= 2(x – 2)2 – 1
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Examination-style question
f(x) can be written as f(x) = 2(x – 2)2 – 1
b)
From this we can see that the minimum value of f(x) is –1.
This occurs when x = 2.
2x2 – 8x + 7 = 0
c)
2(x – 2)2 – 1 = 0
2(x – 2)2 = 1
(x – 2)2 =
1
2
x–2=±
1
2
x=2±
x=2
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1
2
or
x=2+
1
2
1
2
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Examination-style question
d) When y = 0, x = 2 –
1
2
or
x=2+
1
2
When x = 0, y = 7
So the graph cuts the coordinate axes at (2 +
(2 – 21 , 0) and (0, 7).
1
2
, 0),
The parabola has a minimum at the point (2, –1).
y
7
–1
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2–
1
2
2+
1
2
x
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C1.2 Algebra and functions 2